Find the currents flowing in the circuit in Figure 5.

*Strategy*

This circuit is sufficiently complex that the currents cannot be found using Ohm’s law and the series-parallel techniques—it is necessary to use Kirchhoff’s rules. Currents have been labeled I1I1 size 12{I rSub { size 8{1} } } {}, I2I2 size 12{I rSub { size 8{2} } } {}, and I3I3 size 12{I rSub { size 8{3} } } {} in the figure and assumptions have been made about their directions. Locations on the diagram have been labeled with letters a through h. In the solution we will apply the junction and loop rules, seeking three independent equations to allow us to solve for the three unknown currents.

*Solution*

We begin by applying Kirchhoff’s first or junction rule at point a. This gives

I1=I2+I3,I1=I2+I3, size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } } {}

(1)since I1I1 size 12{I rSub { size 8{1} } } {} flows into the junction, while I2I2 size 12{I rSub { size 8{2} } } {} and I3I3 size 12{I rSub { size 8{3} } } {} flow out. Applying the junction rule at e produces exactly the same equation, so that no new information is obtained. This is a single equation with three unknowns—three independent equations are needed, and so the loop rule must be applied.

Now we consider the loop abcdea. Going from a to b, we traverse R2R2 size 12{R rSub { size 8{2} } } {} in the same (assumed) direction of the current I2I2 size 12{I rSub { size 8{2} } } {}, and so the change in potential is −I2R2−I2R2 size 12{ - I rSub { size 8{2} } R rSub { size 8{2} } } {}. Then going from b to c, we go from –– to +, so that the change in potential is +emf1+emf1 size 12{+"emf" rSub { size 8{1} } } {}. Traversing the internal resistance r1r1 size 12{r rSub { size 8{1} } } {} from c to d gives −I2r1−I2r1 size 12{ - I rSub { size 8{2} } r rSub { size 8{1} } } {}. Completing the loop by going from d to a again traverses a resistor in the same direction as its current, giving a change in potential of −I1R1−I1R1 size 12{ - I rSub { size 8{1} } R rSub { size 8{1} } } {}.

The loop rule states that the changes in potential sum to zero. Thus,

−I2R2+emf1−I2r1−I1R1=−I2(R2+r1)+emf1−I1R1=0.−I2R2+emf1−I2r1−I1R1=−I2(R2+r1)+emf1−I1R1=0. size 12{ - I rSub { size 8{2} } R rSub { size 8{2} } +"emf" rSub { size 8{1} } - I rSub { size 8{2} } r rSub { size 8{1} } - I rSub { size 8{1} } R rSub { size 8{1} } = - I rSub { size 8{2} } \( R rSub { size 8{2} } +r rSub { size 8{1} } \) +"emf" rSub { size 8{1} } - I rSub { size 8{1} } R rSub { size 8{1} } =0} {}

(2)Substituting values from the circuit diagram for the resistances and emf, and canceling the ampere unit gives

−3I2+18−6I1=0.−3I2+18−6I1=0. size 12{ - 3I rSub { size 8{2} } +"18" - 6I rSub { size 8{1} } =0} {}

(3)Now applying the loop rule to aefgha (we could have chosen abcdefgha as well) similarly gives

+I1R1+I3R3+I3r2−emf2= +I1R1+I3R3+r2−emf2=0.+I1R1+I3R3+I3r2−emf2= +I1R1+I3R3+r2−emf2=0. size 12{+I rSub { size 8{1} } R rSub { size 8{1} } +I rSub { size 8{3} } R rSub { size 8{3} } +I rSub { size 8{3} } r rSub { size 8{2} } - "emf" rSub { size 8{2} } "=+"I rSub { size 8{1} } R rSub { size 8{1} } +I rSub { size 8{3} } left (R rSub { size 8{3} } +r rSub { size 8{2} } right ) - "emf" rSub { size 8{2} } =0} {}

(4)Note that the signs are reversed compared with the other loop, because elements are traversed in the opposite direction. With values entered, this becomes

+6I1+2I3−45=0.+6I1+2I3−45=0. size 12{+6I rSub { size 8{1} } +2I rSub { size 8{3} } - "45"=0} {}

(5)These three equations are sufficient to solve for the three unknown currents. First, solve the second equation for I2I2 size 12{I rSub { size 8{2} } } {}:

I2=6−2I1.I2=6−2I1. size 12{I rSub { size 8{2} } =6 - 2I rSub { size 8{1} } } {}

(6)Now solve the third equation for I3I3 size 12{I rSub { size 8{3} } } {}:

I3=22.5−3I1.I3=22.5−3I1. size 12{I rSub { size 8{3} } ="22" "." 5 - 3I rSub { size 8{1} } } {}

(7)Substituting these two new equations into the first one allows us to find a value for I1I1 size 12{I rSub { size 8{1} } } {}:

I1=I2+I3=(6−2I1)+(22.5−
3I1)=28.5−
5I1.I1=I2+I3=(6−2I1)+(22.5−
3I1)=28.5−
5I1. size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } = \( 6 - 2I rSub { size 8{1} } \) + \( "22" "." 5 - 3I rSub { size 8{1} } \) ="28" "." 5 - 5I rSub { size 8{1} } } {}

(8)Combining terms gives

6I1=28.5, and6I1=28.5, and size 12{6I rSub { size 8{1} } ="28" "." 5} {}

(9)I1=4.75 A.I1=4.75 A. size 12{I rSub { size 8{1} } =4 "." "75"" A"} {}

(10)Substituting this value for I1I1 size 12{I rSub { size 8{1} } } {} back into the fourth equation gives

I
2
=
6
−
2I
1
=
6
−
9.50
I
2
=
6
−
2I
1
=
6
−
9.50
size 12{I rSub { size 8{2} } =6 - 2I rSub { size 8{1} } =6 - 9 "." "50"} {}

(11)I2=−3.50 A.I2=−3.50 A. size 12{I rSub { size 8{2} } = - 3 "." "50"" A"} {}

(12)The minus sign means I2I2 size 12{I rSub { size 8{2} } } {} flows in the direction opposite to that assumed in Figure 5.

Finally, substituting the value for I1I1 size 12{I rSub { size 8{1} } } {} into the fifth equation gives

I
3
=
22.5
−
3I
1
=
22.5
−
14
.
25
I
3
=
22.5
−
3I
1
=
22.5
−
14
.
25
size 12{I rSub { size 8{3} } ="22" "." 5 - 3I rSub { size 8{1} } ="22" "." 5 - "14" "." "25"} {}

(13)I3=8.25 A.I3=8.25 A. size 12{I rSub { size 8{3} } =8 "." "25"" A"} {}

(14)*Discussion*

Just as a check, we note that indeed I1=I2+I3I1=I2+I3 size 12{I rSub { size 8{1} } =I rSub { size 8{2} } +I rSub { size 8{3} } } {}. The results could also have been checked by entering all of the values into the equation for the abcdefgha loop.

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